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# The difference 942 − 249 is a positive multiple of 7. If a, b, and c a

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Senior Manager
Joined: 04 Sep 2017
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Sep 2019, 13:59
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43% (02:22) correct 57% (02:27) wrong based on 274 sessions

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The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

A. 142
B. 71
C. 99
D. 20
E. 18

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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Sep 2019, 17:07
13
9
The difference between the two digits (abc − cba) = $$100a+10b+c - 100c-10b-a = 99a-99c = 99(a-c)$$
As 99 is not divisible by 7, then $$(a-c)$$ must be divisible by 7 (and b has no effect on the overall outcome)

$$(a-c)$$ are only 2 possible values: $$(9-2)$$ and $$(8-1)$$ ;as a,b,c are non zero; and a must be > c to keep the positive sign.
b has $$9$$ possible values (all except 0)

So the total number of possible outcomes = $$2*9 = 18$$
E
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Sep 2019, 22:38
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gmatt1476 wrote:
The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01

The numbers are 100a + 10b +c and 100c + 10b + a.
After taking the difference, we are left with 99(a-c).
Now for this number to be a multiple of 7, a-c = 7 or 0.
for (a-c) = 7, there are two possibilities ---> (9,2) and (8,1)
and for a-c = 0, a = c, so there are 9 possibilities that are 1 to 9.

The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case.

There are a total of 11 cases so the total number of numbers possible are 9*11 = 99.

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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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22 Sep 2019, 01:32
given
100a+10b+c-100c-10b-a ; 99(a-c)
for 99(a-c) to be divisible (a-c) has to be divisible by 7 and a,b,c are all single digit non zero ;
only possibility of (a-c) = 7 ; ( 9-2) & (8-1) ; 2 choices of a & c ; for b which is middle term we have 9 options ; so total possible values; 9*2 ; 18
IMO E

gmatt1476 wrote:
The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01
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Joined: 03 Jun 2019
Posts: 1950
Location: India
GMAT 1: 690 Q50 V34
Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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26 Sep 2019, 08:58
2
gmatt1476 wrote:
The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01

Given: The difference 942 − 249 is a positive multiple of 7.

Asked: If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

(100a + 10b + c) - (100c + 10b + a) = 99(a-c)
Since 99 is not a multiple of 7, (a-c) should be a multiple of 7.

There are 2 possibilities = {(8,1),(9,2)}
There are 9 possibilities for b

Total such numbers = 2*9 = 18

IMO E
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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26 Sep 2019, 22:10
One question on this - If a, b and c are given as different alphabets, should we consider them as distinct or not?
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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26 Sep 2019, 22:24
Kanika3agg wrote:
One question on this - If a, b and c are given as different alphabets, should we consider them as distinct or not?

No, You can't consider them distinct without being given that they are different/distinct integers or something such as a≠b≠c or a>b>c
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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26 Sep 2019, 22:56
MahmoudFawzy - Thank you!
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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11 Oct 2019, 07:28
Brilliant question. Can someone please post questions like this one ?
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GMAT 1: 750 Q50 V42
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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04 Nov 2019, 23:32
uchihaitachi wrote:
gmatt1476 wrote:
The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?

A. 142
B. 71
C. 99
D. 20
E. 18

PS01661.01

The numbers are 100a + 10b +c and 100c + 10b + a.
After taking the difference, we are left with 99(a-c).
Now for this number to be a multiple of 7, a-c = 7 or 0.
for (a-c) = 7, there are two possibilities ---> (9,2) and (8,1)
and for a-c = 0, a = c, so there are 9 possibilities that are 1 to 9.

The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case.

There are a total of 11 cases so the total number of numbers possible are 9*11 = 99.

By this method, the number 252, 343, 515 etc. all qualify to represent 'abc'.
Let's take 252 as abc.
 cba = 252.
 abc - cba = 0.
But the question wants the difference to be a positive multiple of 7.
Now, 0 is neither positive nor negative.
Thus, when abc = 252, then abc – cba is not a positive multiple of 7.
 The condition a = c is not valid.
Only (a,c) = (9,2) and (8,1) is possible.
In both cases b can take 9 possible values as b ≠ 0.
 2 x 9 = 18 possible choices.
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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14 Dec 2019, 19:11
Can we solve the problem this way?

942 = 3^2 2^2 2^1

Total Factors = (2+1) * (2+1) * (1+1) = 18
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Jan 2020, 02:57
Notice something cool in this question, for abc - cba to be divisible by 7, a-c must equal 7 (you can see that from the question stem) if you don't have a strong number sense, using algebra you could reach the same conclusion.

(1) a,c,b are non-zero digits and they are also different
(2) a-c will equal 7 in two different cases a=9, c=2 and a=8, c=1. In both cases b doesn't affect the divisibility of abc-cba by 7, so it can therefore take any value from 1-9 (that is 7 values in both cases since b cannot equal a or c)

Case 1: a (1 value), b (7 values), and c (1 value) = 9
Case 2: a (1 value), b (7 values), and c (1 value) = 9

Total possible numbers that satisfy the above restrictions: 9+9=18
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Re: The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Jan 2020, 05:43
eduardolarrranaga wrote:
Can we solve the problem this way?

942 = 3^2 2^2 2^1

Total Factors = (2+1) * (2+1) * (1+1) = 18

942 is given as an example to show what the question means. The number of factors of 942 has nothing to do with the solution.
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a  [#permalink]

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21 Jan 2020, 07:00
The difference of a 3-digit number and its reverse is always divisible by 99.

Now, 99 does not have a factor of 7 in it.

Expressing the 3-digit number as xyz and its reverse as zyx, we have their difference to be 99 (x-z) { because xyz can also be written as 100x + 10y + z and zyx as 100z + 10y + x}. So, the (x-z) has to have a factor of 7 if the difference has to be a multiple of 7 as the question says.

Considering that all the digits involved here are non-zero, (x-z) can be a factor of 7 in only two ways i.e. x=9 and z = 2 AND x=8 and z=1.
For each of these cases, y can be dealt in 9 ways i.e. y can be any digit from 1 to 9. Therefore, we have a total of 2*9 = 18 numbers.

The correct answer option is E.

Hope that helps!
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The difference 942 − 249 is a positive multiple of 7. If a, b, and c a   [#permalink] 21 Jan 2020, 07:00
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